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Some silly modeling question

by marmot
Tags: modeling, silly
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Feb26-09, 11:00 PM
P: 53
so my bro ahd an algebra 2 midterm and a question was this:

) A printer costs $35,000 (how old are these questions lol) but it depreciates 5% a year. What is the value by the 4th year?

ok so being an overtly complex person i tried to model a differential equation of this just for the kicks.

At first I thought


where P is price

and the solution is Po*exp(-0.5t) where Po=35000

however this is wrong. So I assumed I did not know r from dP/dt=-rP and worked the problem by finding the initial values.

so the solution gives me P=exp(-.051293t)Po which is correct.

I dont grasp intuitively the answer. why is dP/dt=.051293P when the problem says it goes down 0.5 each year so I assume P changes over time by -0.05P per year?
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Feb26-09, 11:32 PM
P: 287
The correct equation is, of course.

P = P_0 * (0.95)^t.

Another way of writing this is:

P = P_0 * exp(ln(0.95)*t).

As you probably have already guessed, ln(0.95) is the number you need.

ln(0.95) = -.051293, approximately.
Feb26-09, 11:36 PM
P: 53
yeah, thats true. i already knew that but, why isnt dP/dt=-0.5P!!! it does not make sense in my mind. it seems that the changing rate of P is -0.5 P per year

Feb27-09, 07:30 AM
P: 287
Some silly modeling question

Almost certainly because 0.5 isn't the instantaneous rate, but the "average" yearly rate.

What you should say is that the integral from 0 to 1 of e^r equals -0.05...

then e^r = 0.95 and r = ln(0.95).

In fact, that's exactly it. You can't substitute the average (or cumulative, to put it in better terms) rate for the instantaneous rate.
Feb27-09, 01:51 PM
P: 53
that makes a lot of sense! because i can still integrate from 0 to .1 and that wouldnt obviously be right.

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